$\overline{BC} = 10$ $\overline{AC} = {?}$ $A$ $C$ $B$ $?$ $10$ $ \sin( \angle ABC ) = \frac{3\sqrt{109} }{109}, \cos( \angle ABC ) = \frac{10\sqrt{109} }{109}, \tan( \angle ABC ) = \dfrac{3}{10}$
Solution: $\overline{AC}$ is the opposite to $\angle ABC$ $\overline{BC}$ is adjacent to $\angle ABC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle ABC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{AC}}{\overline{BC}}= \frac{\overline{AC}}{10} $ $ \overline{AC}=10 \cdot \tan( \angle ABC ) = 10 \cdot \dfrac{3}{10} = 3$